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So assume we have a sequence of absolutely continuous functions $g_n$ on $[0,1]$ such that $|g_n(0)|\leq 1$ and $|g'(x)|\leq 1$ for Lebesgue almost everywhere. Show that there is a subsequence that converges uniformly to a Lipschitz function.

This is what I have so far. By Lebesgue's Fundamental Theorem of Calculus we have $g_n(x)=g_n(0)+\int_0^xG_n(t)dt$ for some Lebesgue integrable function $G_n$. Since $g_n(0)$ is bounded we can assume that it converges by passing to a subsequence. Thus, it suffices to show that $G_n$ has a subsequence that converges $L^1$ to some integrable function $G$, then uniform convergence is clear since

$$\begin{align} |g_n(x)-g(x)|\leq |g_n(0)-g(0)|+\int_0^1|G_n(t)-G(t)|dt \end{align}$$

The fact this function is Lipschitz is because $g(x)$ is differentiable. So By the mean value theorem and the fact that $|g_n'(x)|\leq 1$ we have $|g'(x)|\leq 1$ and

$$|g(x)-g(y)|\leq |x-y|$$

So my question is how to establish that there is subsequence of $G_n$ that converges in $L^1$. There's probably a theorem for this, but I don't recall any one in particular, so some help here would be helpful.

EDIT: Here is a modified proof using Arzela Ascoli theorem. By Lebesgue's Fundamental Theorem of Calculus we have $g_n(x)=g_n(0)+\int_0^xG_n(t)dt$ for some Lebesgue integrable function $G_n$.In our case $G_n=g_n'$ a.e. From this we show that $g_n$ are uniformly bounded using the fact that $|g_n(0)|\leq 1$ and $|g_n'(x)|\leq 1$

$$\begin{align} |g_n(x)|&\leq|g_n(0)|+\int_0^x|G_n(t)|dt\\ &\leq1+\int_0^1dt\\ &=2\\\ \end{align}$$

To show equicontinuity we note that $$\begin{align} |g_n(x)-g_n(y)|&\leq\int_y^x|G_n(t)|dt\\ &\leq|x-y|\\ \end{align}$$

Thus, there exists a subsequence that converges uniformly to a function $g$. Now we use the fact that given a metric $d$ and sequences $x_n,y_n$ that converges to $x,y$ respectively we have $d(x_n,y_n)\rightarrow d(x,y)$. We've shown that $g_n$ are Lipschitz. So passing to the limit we have $|g(x)-g(y)|\leq |x-y|$.

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  • $\begingroup$ Do you know about the Arzèla Ascoli theorem? If not, familiarize yourself with it... it's pretty important! $\endgroup$ – Shalop Sep 2 '17 at 21:47
  • $\begingroup$ Oh, so all I need to show is that the sequence is equicontinuous since our functions are bounded by 2. $\endgroup$ – thegamer Sep 2 '17 at 21:48
  • $\begingroup$ What else do you know about $G_n$? $\endgroup$ – user357980 Sep 2 '17 at 21:49

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